Primitive of Root of 2 a x minus x squared

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \sqrt {2 a x - x^2} \rd x = \frac {\paren {x - a} } 2 \sqrt {2 a x - x^2} + \frac {a^2} 2 \arcsin \frac {x - a} a + C$

where $C$ is an arbitrary constant.


Proof

Let $u := x - a$.

Then:

$\dfrac {\d u} {\d x} = 1$

and:

$x = u + a$


Then:

\(\ds 2 a x - x^2\) \(=\) \(\ds 2 a \paren {u + a} - \paren {u + a}^2\)
\(\ds \) \(=\) \(\ds 2 a u + 2 a^2 - u^2 - 2 a u - a^2\)
\(\ds \) \(=\) \(\ds a^2 - u^2\)


and we have:

\(\ds \int \sqrt {2 a x - x^2} \rd x\) \(=\) \(\ds \int \sqrt {a^2 - u^2} \rd u\)
\(\ds \) \(=\) \(\ds \frac {u \sqrt {a^2 - u^2} } 2 + \frac {a^2} 2 \arcsin \frac u a + C\) Primitive of $\sqrt {a^2 - u^2}$: Arcsine Form
\(\ds \) \(=\) \(\ds \frac {\paren {x - a} } 2 \sqrt {2 a x - x^2} + \frac {a^2} 2 \arcsin \frac {x - a} a + C\) substituting for $u$ and simplifying

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Root_of_2_a_x_minus_x_squared&oldid=675147"